The voltage drop across a forward biased diode is 0.7b V. In the following circuit, the voltages across the 10Ω resistance in series with the diode and 20Ω resistance are:

3.58 V, 4.28 V

Let the currents through the 20Ω (parallel) and 10Ω (in series with the diode) be $I_1$ and $I_2$ respectively.

Applying Kirchhoff’s second law for closed loop ABEFA, we get

$20I_1 +10(I_1 + I_2 ) −10 = 0$ ......(i)

Applying Kirchhoff’s second law for closed loop BCDEB, we get

$0.7 +10I_2 − 20I_1 = 0$ .......(ii)

Solving (i) and (ii), we get

$I_1 = 0.214A$ and $I_2 = 0.358A$

Thus, voltage across the 10Ω resistance in series with the diode = $0.358A×10Ω = 3.58V$

And voltage across the 20Ω resistance = $0.214A×20Ω = 4.28V$